Analytical Chem istry - DePauw University

Chapter 9 Titrimetric Methods449a pH of 4.5. Identify the sources of alkalinity and their concentrations inmilligrams per liter.So l u t i o nBecause the volume of titrant to reach a pH of 4.5 is more than twice thatneeded to reach a pH of 8.3, we know, from Table 9.6, that the sample’salkalinity is controlled by CO 3 2– and HCO 3 – . Titrating to a pH of 8.3neutralizes CO 3 2– to HCO 3–2− − −CO ( aq) + HCl( aq) → HCO ( aq) + Cl ( aq)3but there is no reaction between the titrant and HCO 3 – (see Figure 9.19).The concentration of CO 3 2– in the sample, therefore, is0. 02812 MHCl× 0.01867 LHCl×5.250×100.1000 LmolCO1 molCOmol HCl2−33= 5.250× 10 −4 molCO 2−360.01 gCO 1000 mg×× = 315.1 mg/L2− mol CO g− 4 2 − 2 −3 3Titrating to a pH of 4.5 neutralizes CO 3 2– to H 2 CO 3 , and HCO 3 – toH 2 CO 3 (see Figures 9.19).2−−CO ( aq) + 2HCl( aq) → H CO ( aq) + 2Cl( aq)332 3−−HCO ( aq) + HCl( aq) → HCO ( aq) + Cl ( aq)23 2 3Because we know how many moles of CO 3 2– are in the sample, we cancalculate the volume of HCl it consumes.−4 2−2 molHCl5. 250× 10 molCO × ×32−molCO1 L HCl 1000 mL× =0.02812 molHCl L337.34This leaves 48.12 mL - 37.34 mL, or 10.78 mL of HCl to react withHCO 3 – . The amount of HCO 3 – in the sample is3.031×100. 02812 MHCl× 0.01078 LHCl×0.1000 LmolHCO1 molHCOmol HCl−3mL= 3.031× 10 −4molHCO −361.02 gHCO mg×mol HCO× 1000− g=−4− −3 33185.0mg/L